Find the value of x that solves the equation ln(x-5)-ln 2=0. Answer:

Combine logarithms: We need to combine the logarithms on the left side of the equation using the property of logarithms that states ln(a) - ln(b) = ln(a/b). So, we rewrite the equation as ln((x-5)/2) = 0.Get rid of ln: To solve for x, we need to get rid of the natural logarithm. We can do this by raising both sides of the equation as powers of e, because e to the power of ln(a) is just a. So, we have e^(ln((x-5)/2)) = e^0.Raise as powers of e: Since e^0 is 1, we now have (x-5)/2 = 1.Simplify equation: To solve for x, we multiply both sides of the equation by 2 to get x-5 = 2.Multiply by 2: Finally, we add 5 to both sides of the equation to solve for x, which gives us x = 2 + 5.Add 5: So, x = 7 is the solution to the equation.

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Q. Find the value of

x

that solves the equation

\ln (x-5)-\ln 2=0

\newline

Answer:

Combine logarithms: We need to combine the logarithms on the left side of the equation using the property of logarithms that states $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ . So, we rewrite the equation as $\ln\left(\frac{x-5}{2}\right) = 0$ .
Get rid of ln: To solve for $x$ , we need to get rid of the natural logarithm. We can do this by raising both sides of the equation as powers of $e$ , because $e$ to the power of $\ln(a)$ is just $a$ . So, we have $e^{\ln((x-5)/2)} = e^0$ .
Raise as powers of e: Since $e^0$ is $1$ , we now have $(x-5)/2 = 1$ .
Simplify equation: To solve for $x$ , we multiply both sides of the equation by $2$ to get $x-5 = 2$ .
Multiply by $2$ : Finally, we add $5$ to both sides of the equation to solve for $x$ , which gives us $x = 2 + 5$ .
Add $5$ : So, $x = 7$ is the solution to the equation.